$\dfrac{d}{dx}[3\cos(x)+4x^2]=$
The expression to differentiate includes $\cos(x)$. Remember that the derivative of $\cos(x)$ is $-\sin(x)$. Put another way, $\dfrac{d}{dx}[\cos(x)]=-\sin(x)$. $\begin{aligned} &\phantom{=}\dfrac{d}{dx}[3\cos(x)+4x^2] \\\\ &=3\dfrac{d}{dx}[\cos(x)]+4\dfrac{d}{dx}(x^2) \\\\ &=3(-\sin(x))+4\cdot2x \\\\ &=-3\sin(x)+8x \end{aligned}$ In conclusion, $\dfrac{d}{dx}[3\cos(x)+4x^2]=-3\sin(x)+8x$